Year 1 of the physics course of the University of Oxford (Prelims)

Four core written papers, a short-option paper and laboratory work.

Core first-year classical mechanics and relativity.

Core first-year paper on differential equations and waves.

Core first-year paper including electromagnetism.

Core first-year mathematical methods.

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

All questions and inline Roman subparts are covered. The solutions distinguish signed from unsigned distances, branches of complex powers, and the exceptional zero-frequency differential equation.

Let $\Delta =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})\ne 0$. Taking the scalar product with $\mathbf{b}\times \mathbf{c}$ in the first expansion kills its other two terms. Taking the scalar product with $\mathbf{a}$ in the second expansion does the same. Thus the two requested coordinates areThe scalar triple product condition guarantees linear independence and the existence of these reciprocal bases. The second coordinate is not obtained by merely normalizing the first basis vector: the two expansions use different bases.

On an interval with $x\ne 0$, the zero function is one solution. For a nonzero solution, set $u=y^{-2}$. Dividing the equation by two and differentiating $u$ gives the linear ordinary differential equationIts integrating factor is $x^{-2}$, so $(u/x^2{)}^{\prime }=-x^{-5}$ and $u=1/(4x^2)+C{x}^2$. Absorbing the factor four into the integration constant givesFor real solutions, choose a connected interval avoiding $x=0$ on which $1+K{x}^4>0$, with a fixed sign. The special straight solutions $y=\pm 2x$ are already included at $K=0$. The separate zero solution would be lost by the division used in the substitution.

Using the complex-angle identity,Since $\cosh y>0$, the real part can vanish only when $x=\pi /2+k\pi$. Then the imaginary part requires $\sinh y=(-1{)}^{k+1}$. Therefore all roots areSubstitution gives $\cos z=i$ for every integer $k$, and the positivity of $\cosh y$ shows that no other real parts were missed. The imaginary offsets are the two values $\pm \mathrm{arsinh}1$.

Set $v=\dot{z}$. The second-order position equation becomes the first-order linear ordinary differential equation $\dot{v}+\gamma v=-g$. For $\gamma >0$, multiply by the integrating factor $e^{\gamma t}$ and impose $v(0)=v_0$:The limiting terminal velocity is $-g/\gamma$, consistent with upward-positive position and downward gravity. The continuous zero-drag limit is $v_0-gt$; one must take that limit of the whole expression rather than set $\gamma =0$ in its two singular-looking terms separately.

Multiplication by $A$ rotates the $(x,y)$ components by $\theta$ counterclockwise when viewed from positive $z$ toward the origin, leaving $z$ unchanged. It is a proper rotation about the $z$ axis, with determinant $+1$. Multiplication by $B$ sends $(x,y,z)$ to $(x,y,-z)$: it is a reflection in the $xy$ plane, with determinant $-1$. Neither operation changes a vector's Euclidean norm.

Directly multiplying the columns' scalar products gives $C^{T}C=I$, and direct evaluation gives $\det C=1$. Thus the operation is a proper rotation. Its trace is zero. A three-dimensional proper rotation has eigenvalues $1,e^{i\phi },e^{-i\phi }$, hence $\mathrm{tr}C=1+2\cos \phi$. ThereforeThis determines the usual unsigned angle in $[0,\pi ]$ without changing coordinates. The axis can be checked, if desired, from $C(0,1,-1{)}^T=(0,1,-1{)}^T$. The determinant test must precede the trace formula: an improper rotation instead has trace $-1+2\cos \phi$.

Let $a=R/(2L)$ and ${\omega }_d=\sqrt{1/(LC)-a^2}>0$. The roots of the characteristic polynomial are $-a\pm i{\omega }_d$, so the real underdamped solution is $Q=e^{-at}(A\cos {\omega }_{d}t+B\sin {\omega }_{d}t)$. The initial charge gives $A=Q_0$ and the zero initial current gives $B=a{Q}_0/{\omega }_d$. HenceThis solves the homogeneous RLC circuit equation supplied in the question. A nonzero applied voltage would appear as a forcing term and require a particular solution; it is not present in that equation.

The additive identity for the shifted operation must satisfy $\mathbf{x}+0_{\oplus }-\mathbf{a}=\mathbf{x}$, giving $0_{\oplus }=\mathbf{a}$. The additive inverse must satisfy $\mathbf{x}+{\mathbf{x}}^{\prime }-\mathbf{a}=\mathbf{a}$, givingThe notation distinguishes the new additive identity from the ordinary zero vector. This is a translated abelian group structure on the same underlying set.

The translated vector-space structure follows by translating the new origin to the ordinary one using the bijection $T(\mathbf{x})=\mathbf{x}-\mathbf{a}$. Transporting ordinary scalar multiplication through $T$ givesThen $T(\mathbf{x}\oplus \mathbf{y})=T(\mathbf{x})+T(\mathbf{y})$ and $T(k\odot \mathbf{x})=kT(\mathbf{x})$, so the transported operations satisfy all vector space axioms. In particular, an explicit distributive check isThe other distributive law follows in the same coordinates, or directly by expanding $(k+l)\odot \mathbf{x}$.

For $c>0$, this is the sphere centered at $\mathbf{a}$ with radius $c$. If $c=0$, its locus is the singleton $\{\mathbf{a}\}$; if $c<0$, it is empty, because a Euclidean norm cannot be negative. The equation describes the surface, not the enclosed ball.

Assume $\mathbf{a}\ne 0$ and allow all $\mathbf{R}\ne 0$. Put $\hat{\mathbf{a}}=\mathbf{a}/|\mathbf{a}|$ and $\mathbf{u}=\mathbf{R}/|\mathbf{R}|$. Every resulting vector is perpendicular to $\mathbf{a}$ and has $|\mathbf{r}|=|c|\sin \theta \le |c|$. Conversely, for $c\ne 0$ and any $\mathbf{q}=\mathbf{r}/c$ perpendicular to $\hat{\mathbf{a}}$ with $|\mathbf{q}|\le 1$, chooseThe vector triple product gives $\mathbf{u}\times \hat{\mathbf{a}}=\mathbf{q}$, and $|\mathbf{u}|=1$. Thus every point of the closed disk centered at the origin, perpendicular to $\mathbf{a}$, of radius $|c|$ is attained. The boundary circle alone would omit allowed directions of $\mathbf{R}$. For $c=0$ the locus is the origin; for $\mathbf{a}=0$ the original expression is undefined.

A vector from the point on $L_2$ to the point on $L_1$ isChoosing the opposite direction negates this vector and does not change its norm or the shortest distance.

For nonparallel direction vectors, let $\mathbf{n}=(\mathbf{b}\times \mathbf{d})/|\mathbf{b}\times \mathbf{d}|$. Every joining vector has the same normal component, $\mathbf{R}\cdot \mathbf{n}=(\mathbf{a}-\mathbf{c})\cdot \mathbf{n}$. Its norm is therefore at least the absolute value of that component. Minimizing $|\mathbf{R}{|}^2$ makes it perpendicular to both directions, and hence parallel to $\mathbf{n}$, attaining the bound:The expression printed without absolute values is an oriented separation, not a nonnegative distance. Reversing either line's parametrization can flip its sign. For parallel nonzero directions, this normal is undefined and the distance is instead $|(\mathbf{a}-\mathbf{c})\times \mathbf{b}|/|\mathbf{b}|$.

Write $\mathbf{w}=\mathbf{a}-\mathbf{c}$, $B=\mathbf{b}\cdot \mathbf{b}$, $D=\mathbf{d}\cdot \mathbf{d}$ and $E=\mathbf{b}\cdot \mathbf{d}$. Differentiating $|\mathbf{w}+\lambda \mathbf{b}-\mu \mathbf{d}{|}^2$ with respect to the two parameters givesFor nonparallel directions, $BD-E^2=|\mathbf{b}\times \mathbf{d}{|}^2>0$. Solving this linear system yieldsThe quadratic being minimized is strictly convex in the two parameters under this nonparallel assumption, so the critical point gives the unique pair of closest points. Parallel lines do not select a unique value of $\lambda$.

Parametrize the first line by $(x,y,z)=(3+t,4-2t,1+3t)$ and the second by $(2+\alpha s,-1+s,5+2s)$. Equality of the $y,z$ components gives $s=5-2t$ and $3t-2s=4$, hence $t=2$, $s=1$. The $x$ components then require $5=2+\alpha$. ThusThe two independent component equations fixed the parameters before the remaining component determined $\alpha$, so no extra intersection case was discarded.

The right-hand side has modulus two and argument $\pi /3$. Its fifth roots therefore haveThey are the vertices of a regular pentagon on the circle of radius $2^{1/5}$, beginning at angle $\pi /15$. These five roots are distinct, and a degree-five polynomial has no further complex roots.

The base is $2e^{i\pi /4}$. Its complex logarithms are $\ln 2+i(\pi /4+2\pi k)$, $k\in \mathbb{Z}$. Exponentiating $i$ times each logarithm gives all values of the multivalued power:The values form a discrete geometric sequence along the Argand ray of angle $\ln 2$. Consecutive radii differ by $e^{-2\pi }$. The origin is an accumulation point but is not itself a value. If the power is instead defined using the principal logarithm, only $k=0$ is retained. The phrase “all values” requires this branch distinction.

For $z\ne -1$, multiply numerator and denominator by $\overline{z}+1$. The real part isThus its zero locus isIt is the unit circle with the point $-1$ removed. That point cannot be restored by the zero numerator obtained after clearing a denominator: the original expression is undefined there.

Use the imaginary part of the finite complex sum:The identity $1+e^{i\theta }=2e^{i\theta /2}\cos (\theta /2)$ then givesHere $N$ is a nonnegative integer. Keeping the signed cosine raised to the integer power makes the result valid even when $\cos (\theta /2)<0$; inserting an absolute value would lose the sign for odd $N$.

For unitary matrices, $U^{†}U=I$ and $V^{†}V=I$. The adjoint matrix reverses the order of multiplication, soThus the product is unitary. Likewise $(UV)(UV{)}^{†}=I$, or this follows from invertibility in finite dimension.

Take the adjoint matrix of $M^{†}{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i$ to obtain ${\mathbf{v}}_i^{†}M={\lambda }_i^{\ast }{\mathbf{v}}_i^{†}$. Evaluating ${\mathbf{v}}_i^{†}M{\mathbf{u}}_j$ using this equation and $M{\mathbf{u}}_j={\lambda }_j{\mathbf{u}}_j$ givesHence $\boxed{{\mathbf{v}}_i^{†}{\mathbf{u}}_j=0\text{ if }{\lambda }_i^{\ast }\ne {\lambda }_j}$. The proof does not assume that $M$ is a normal matrix or diagonalizable.

In general the spectrum of an adjoint matrix is the complex conjugate of the original spectrum, not necessarily the same set. The equality of the two sets in the question is an additional assumption, possible for a conjugation-invariant spectrum; the displayed orthogonality proof only uses the two stated eigenvalue equations.

A column of coordinates transforms as ${\mathbf{x}}^{\prime }=U^{†}\mathbf{x}$, because the columns of $U$ are the new orthonormal basis vectors expressed in the old coordinates. A symmetric unitary choice isIts first and third columns are orthogonal eigenvectors of the $xz$ block, with eigenvalues $-1$ and $+1$ respectively; its second column is the $y$ axis with eigenvalue $3$. Direct multiplication checks $U^{†}U=I$, while the displayed entries give $U^T=U$. The symmetry here means transpose symmetry, not Hermitian symmetry.

Separation of variables gives $\ln |y|=\sqrt{1+x^2}+C$ for a nonzero solution. Including the zero solution in the arbitrary real amplitude givesDifferentiating the exponent verifies the equation. This form keeps the zero solution that a division by $y$ could otherwise conceal.

On the component $x>0$ containing the boundary point, divide by $2x$ and use integrating factor $x^{-1/2}$:Integration gives $y=x^3/5+C\sqrt{x}$, and $y(1)=1$ fixes $C=4/5$. ThereforeAlthough the function has a continuous limit at zero, its derivative diverges there; the solution with this boundary datum does not extend as a differentiable solution through the singular point of the equation.

The exponential substitution removing a constant first derivative, $y=e^{x}u$, reduces the equation to $u^{\prime \prime }+{\beta }^{2}u=1+\cos x$. For $\beta \ne 0$ and ${\beta }^2\ne 1$, solve the homogeneous harmonic equation and use a constant plus a cosine for the particular solution:There is a further exceptional case inside the printed condition ${\beta }^2\ne 1$: $\beta =0$ must be treated separately. It gives $u^{\prime \prime }=1+\cos x$, henceThis retains the constant forcing's resonance with the zero-frequency homogeneous equation. Substitution verifies both branches.

For ${\beta }^2=1$, the reduced equation is $u^{\prime \prime }+u=1+\cos x$. The constant forcing has particular solution one, while the resonant cosine requires $u_p=(x/2)\sin x$. ThusThe secular factor $x$ is the signature of resonance. It must replace the divergent nonresonant coefficient $1/({\beta }^2-1)$; substituting ${\beta }^2=1$ directly into that coefficient is invalid.

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

The polar form is $z=e^{3\pi i/4}$. Interpreting the fractional complex exponentiation as all its values, a candidate $w$ satisfies $w^3=z^2=-i$. The three roots of a complex number areThey have arguments $\pi /2$, $7\pi /6$ and $11\pi /6$, respectively. Since all have modulus one, their squared distances from $z$ in the complex plane are $|w-z{|}^2=2-2\cos (\arg w-3\pi /4)$, givingThe last is strictly largest, so the required root of a complex number isRestricting complex exponentiation to its principal value would give only $i$; the request to choose among roots requires all three values.

Multiply numerator and denominator by the denominator's complex conjugate:Thus the real part and imaginary part are

The polar form is $1+i\sqrt{3}=2e^{i\pi /3}$. By De Moivre's theorem,Therefore the real part and imaginary part are

Using Euler's formula, the difference of the two complex exponentials is $2i\sin (\pi /4)=i\sqrt{2}$. HenceThese are the real part and imaginary part, respectively.

The normal vectors are ${\mathbf{n}}_A=(1,2,-4)$ and ${\mathbf{n}}_B=(2,-1,3)$. Their dot product is $-12$, and their norms are $\sqrt{21}$ and $\sqrt{14}$. The usual angle between two planes is the smaller, unoriented angle, soFor the particular signed normal vectors printed above, the angle is the supplement $\pi -\theta \approx 134.4^{\circ }$. Reversing either normal vector exchanges these two angles without changing either plane.

A parallel plane has the same normal vector, so write $C:x+2y-4z=d$. Substitution of the given point gives $d=4+4-4=4$. ThusA displacement between the two planes has a component of magnitude $|5-4|/\Vert {\mathbf{n}}_A\Vert$ along their common unit vector normal. The distance between parallel planes is consequently

The parallelepiped volume is the absolute value of the scalar triple product of its edge vectors, equivalently the absolute determinant of the matrix with those columns. Expansion along the first row givesTherefore

Multiplying a three-dimensional matrix by a real $\lambda$ multiplies its determinant by ${\lambda }^3$, whereas multiplicativity of the determinant gives $\det (A^5)=(\det A{)}^5$. Therefore equality of the two parallelepiped volumes meansBoth real signs are allowed:A negative value reverses the oriented volume while preserving the ordinary volume. If a positive scale factor is intended, select the positive sign.

The characteristic polynomial isThus the eigenvalues are $i$ and $-i$. There is no real diagonalization, because a real diagonal matrix would have real eigenvalues. Over the complex numbers, the two distinct eigenvalues give linearly independent eigenvectors, so the matrix is diagonalizable.

For $\lambda =i$, the first row gives $(1-i)x+2y=0$. Choose $x=2$, giving $y=-1+i$. Its standard complex norm is $\sqrt{|2{|}^2+|-1+i{|}^2}=\sqrt{6}$; the conjugate eigenvector belongs to $-i$. Hence normalized eigenvectors and a suitable change of basis areDirectly, $AP=P\mathrm{diag}(i,-i)$ and $\det P=-2i/3\ne 0$, soNormalization does not assert that these eigenvectors are orthogonal.

An invertible matrix is a square matrix $X$ admitting a matrix inverse $X^{-1}$ with $X{X}^{-1}=X^{-1}X=I$, where $I$ is the identity matrix. Equivalently, its associated linear map is bijective.

Let $X\mathbf{v}=\lambda \mathbf{v}$ with eigenvector $\mathbf{v}\ne 0$. The eigenvalue cannot be zero: otherwise applying $X^{-1}$ gives $\mathbf{v}=0$. Apply $X^{-1}$ to the eigenvector equation:Thus $1/\lambda$ is an eigenvalue of $X^{-1}$ with the same eigenvector.

An exact differential $Mdx+Ndy$ is $d\Phi$ for a scalar potential function $\Phi$, so $M=\partial \Phi /\partial x$ and $N=\partial \Phi /\partial y$. Equality $\partial M/\partial y=\partial N/\partial x$ is necessary, and is sufficient on a simply connected space when these partial derivatives are continuous.

Here, after multiplying the ordinary differential equation by $dx$,On either half-plane $y>0$ or $y<0$, integrate $M$ with respect to $x$:Comparison of $\partial \Phi /\partial y$ with $N$ gives $h^{\prime }(y)=6y$, hence $h(y)=3y^2$ up to a constant. The solution is therefore the family of level setsLocally these define differentiable graphs $y(x)$ wherever $N\ne 0$, by the implicit function theorem. When $N=0$, $M\ne 0$ on this domain, so a finite derivative cannot satisfy $M+N{y}^{\prime }=0$ there; such points are vertical tangents of the implicit curves, rather than additional classical graph solutions.

Differentiation satisfies $D(af+bg)=aD(f)+bD(g)$ for constant scalars $a,b$, so it is a linear map on suitable vector spaces of differentiable functions. An arbitrary set $S$ need not even be a vector space, and an endomorphism interpretation also requires $D(S)\subseteq S$.

For a noninvertible example, take $S=\mathrm{span}\{1,x\}$. Then $D(a+bx)=b$ and the kernel of a linear map is the nonzero linear span of $1$, so $D:S\to S$ is not injective. The matrix in this basis is $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

For an invertible example, take $S=\mathrm{span}\{e^x,e^{2x}\}$. In this basis, $D$ has matrix $\mathrm{diag}(1,2)$, andThus differentiation is not always invertible; the domain and codomain determine the answer.

For an invertible matrix $P$ whose columns give the new basis in old coordinates, the new matrix is $B=P^{-1}AP$. Multiplicativity of the determinant givessince $\det (P^{-1})\det P=1$.

To justify the cyclic property of the trace directly, for two square matrices $R,S$,Apply it with $R=P^{-1}A$ and $S=P$:Thus both the determinant and trace are invariant under change of basis.